wu :: forums
« wu :: forums - Diameter of f(D) »

Welcome, Guest. Please Login or Register.
Mar 28th, 2024, 3:22pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   general
   complex analysis
(Moderators: william wu, Grimbal, Icarus, ThudnBlunder, towr, SMQ, Eigenray)
   Diameter of f(D)
« Previous topic | Next topic »
Pages: 1 2  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Diameter of f(D)  (Read 6673 times)
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Diameter of f(D)  
« on: Dec 3rd, 2005, 5:12pm »
Quote Quote Modify Modify

Suppose f : D -> C is holomorphic, where D is the unit disc.  If
d = supz,w in D |f(z)-f(w)|
is the diameter of f(D), show that
2|f'(0)| < d,
and that equality holds precisely when f is linear.
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #1 on: Dec 9th, 2005, 7:32pm »
Quote Quote Modify Modify

Wow. This is a nice find.
 
Am I allowed to up some hint for a counterexample?
« Last Edit: Dec 9th, 2005, 7:35pm by Michael Dagg » IP Logged

Regards,
Michael Dagg
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Diameter of f(D)  
« Reply #2 on: Dec 10th, 2005, 6:00pm »
Quote Quote Modify Modify

I don't understand what you mean.
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #3 on: Dec 11th, 2005, 7:38am »
Quote Quote Modify Modify

I had a counterexample hint in mind, which is what I meant, but here is a better hint: use a compactness argument to find a (closed) disk of diameter d which contains f(D). If its center is c, then (2/d)(f - c) maps D into  D. Then apply Schwarz.
IP Logged

Regards,
Michael Dagg
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Diameter of f(D)  
« Reply #4 on: Dec 11th, 2005, 1:31pm »
Quote Quote Modify Modify

But if, say, f(D) is an equilateral triangle with diameter (side length) d, then the smallest disk containing it has diameter 2d/sqrt(3) > d.
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #5 on: Dec 19th, 2005, 1:22pm »
Quote Quote Modify Modify

Indeed - your tactful advisory correct.
 
Note that F(z) = f(z) - f(-z) maps D into D, the disk centered at 0 of radius d, and F(0) = 0, so Schwarz is applicable to F/d.
IP Logged

Regards,
Michael Dagg
cain
Guest

Email

Re: Diameter of f(D)  
« Reply #6 on: Dec 21st, 2005, 11:30am »
Quote Quote Modify Modify Remove Remove

How did you think up that F?
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #7 on: Dec 21st, 2005, 12:32pm »
Quote Quote Modify Modify

The short answer is that Eigenray made me think it up.  
 
What can you say about that F? (I see the mapping problem you posted.)
IP Logged

Regards,
Michael Dagg
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Diameter of f(D)  
« Reply #8 on: Dec 21st, 2005, 6:21pm »
Quote Quote Modify Modify

on Dec 19th, 2005, 1:22pm, Michael_Dagg wrote:
Note that F(z) = f(z) - f(-z) maps D into D, the disk centered at 0 of radius d, and F(0) = 0, so Schwarz is applicable to F/d.

That takes care of the odd coefficients, but what about the even ones?  That is, Schwarz tells you that if equality holds, then F(z) = dz, i.e., f(z) = d/2 z + g(z2) for some g.
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #9 on: Dec 29th, 2005, 5:46am »
Quote Quote Modify Modify

I've tried out a few things and haven't come up with the result for equality. It is an interesting problem as it first made me think Polya, Landau and Toeplitz. Counterexample still sticks out at me but I don't see it at the moment.
IP Logged

Regards,
Michael Dagg
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #10 on: Jan 3rd, 2006, 3:38pm »
Quote Quote Modify Modify

Do you have a result for equality?
IP Logged

Regards,
Michael Dagg
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Diameter of f(D)  
« Reply #11 on: Jan 3rd, 2006, 8:21pm »
Quote Quote Modify Modify

Alas, no.  The inequality is given as an exercise in Stein & Shakarchi's Complex Analysis.  It then says "moreover, it can be shown that equality holds precisely when f is linear," but no reference is given.
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #12 on: Jan 4th, 2006, 7:26am »
Quote Quote Modify Modify

My suspecion gave light: after a closer search I did find the problem in Polya and Szego's PROBLEMS AND THEOREMS IN ANALYSIS,III.239. They reference a 1907 paper of Landau and Toeplitz in a now defunct journal that our library does not have (Arch.der Math.und Physik,ser.3,vol.11, pp.302-307), but I suspect it will also be found in Landau's multi-volume COLLECTED WORKS, which (mirabili dictu) our library does have.
 
There's no guarantee that they discuss the case of equality, but Landau was so meticulous that I suspect they did.  
 
Anyway, Polya and Szego's solution for the inequality is very similar to the one I had found, namely, via consideration of the function  f(z)-f(-z)  ;but they do not consider equality.
« Last Edit: Jan 4th, 2006, 7:32am by Michael Dagg » IP Logged

Regards,
Michael Dagg
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)   diampblm.JPG
« Reply #13 on: Jan 23rd, 2006, 2:12pm »
Quote Quote Modify Modify

I have a solution to this problem that I post as a scanned image. The quality is not very good  
but I would welcome its tagging by one of the gladiators here.
 
I introduce the oscillation F(z) = f(z) - f(-z) (third one) subsequent to the first two by Landau-Toeplitz.
It turns out (as you might expect, although the proof is a bit subtle) that all these oscillations are equal, but the proof I  
wrote skirts this matter. Actually, Landau-Toeplitz worked only with functions that are holomorphic in a neighborhood  
of the closed unit disk. It takes more work to relax that to mere holomorphy in the open unit disk, where, e.g., one can no  
longer speak of the oscillation on the unit circle as before.
 
Please alert me to any typos.
IP Logged


Regards,
Michael Dagg
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Diameter of f(D)  
« Reply #14 on: Jan 27th, 2006, 4:58pm »
Quote Quote Modify Modify

Looks good!  Thanks.  For something this long, I think LaTeX is just easier.  Unfortunately,
"This attachment causes the attachments directory to exceed it's [sic] maximum capacity by approximately 58 kilobytes and cannot be uploaded."
So it may be found here.
« Last Edit: Jan 27th, 2006, 5:03pm by Eigenray » IP Logged
Icarus
wu::riddles Moderator
Uberpuzzler
*****



Boldly going where even angels fear to tread.

   


Gender: male
Posts: 4863
Re: Diameter of f(D)  
« Reply #15 on: Jan 28th, 2006, 11:03am »
Quote Quote Modify Modify

I tried to convert this too, but ran into the same problem.
IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #16 on: Jan 31st, 2006, 8:53am »
Quote Quote Modify Modify

That's nice Eigenray. I noticed the attachment limitation also when I tried to put up a png file that has better quality instead of the jpeg.
« Last Edit: Jan 31st, 2006, 8:55am by Michael Dagg » IP Logged

Regards,
Michael Dagg
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #17 on: Feb 1st, 2006, 11:48am »
Quote Quote Modify Modify

This problem is really good food (that is, the thought of it might be addictive) - one idea I want to add, which is hard to resist, and something you likely already know, e.g. :
 
Let d = diam f(D). Fix (a) in D and put W(z) = (z-a)/(1-a bar z).  
 
Then by Schwarz we have
 
|(f(z) - f(a))/W(z)| <= max|((f(w) - f(a))/ W(w)| over {|w| = 1},
 
<= d.
 
Letting z approach (a), this gives
 
|f'(a)|(1 - |a|^2) <= d.
 
Take a=0, then |f'(0)| <= d.
 
This result is evident (albeit contrived differently) from the solution I posted.
 
Similar arguments give the (best possible, as equality holds for conformal automorphisms) inequality
 
|f'(a)| <=  (1 - |f(a)|^2)/(1 - |a|^2)
 
for every (a) in D, and every holomorphic self-map f of D. You'll find this in the famous 1912 Math. Annalen paper of Caratheodory
and maybe even in Schwarz' original papers (and most graduate-level CA textbooks), however, providing  
a construction (e.g., the function W or otherwise) may take some doing.
 
(Previously I mentioned that I had tried out a few things and this is one of such - made complete sense even without knowing W).
« Last Edit: Feb 2nd, 2006, 1:45am by Michael Dagg » IP Logged

Regards,
Michael Dagg
Paul_W
Newbie
*





   
Email

Posts: 4
Re: Diameter of f(D)  
« Reply #18 on: Feb 3rd, 2006, 2:55pm »
Quote Quote Modify Modify

It would certainly be tempting to instead write W(z) = 2 (z-a) / (1 - a bar z) then get 2 |f'(0)| <= d. But, there is a problem with that.
IP Logged
Michael Dagg
Senior Riddler
****






   


Gender: male
Posts: 500
Re: Diameter of f(D)  
« Reply #19 on: Feb 4th, 2006, 12:16pm »
Quote Quote Modify Modify

I think you meant to write   W2(z) = W(z)/2  . But neither are plausible and if one actually understands the problem the use of them would not be considered.
 
Getting inequality by way of some acceptable W is possible but getting equality this same way would require you come up with  f  in some linear form (and then when combined with  W ) applicable to Schwarz.
 
I can't say it is not possible but say it will likely not be easy (e.g., I never could get any useful bounds going this route but I still think about it).
« Last Edit: Feb 4th, 2006, 6:40pm by Michael Dagg » IP Logged

Regards,
Michael Dagg
Paul_W
Newbie
*





   
Email

Posts: 4
Re: Diameter of f(D)  
« Reply #20 on: Feb 6th, 2006, 11:37am »
Quote Quote Modify Modify

The numbering of the theorems (1,2) in hand-written one differs
from the numbering in the pdf file (3,4). Does it matter (because  
the last sentence of the hand-written one references Theorem 1, which  
would be 3 in the pdf) or does the result (5) actually follow from Lemma 1?
IP Logged
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Diameter of f(D)  
« Reply #21 on: Feb 6th, 2006, 2:23pm »
Quote Quote Modify Modify

Whoops... forgot about that.
 
< \newtheorem{lemma}[thm]{Lemma}
---
> \newtheorem{lemma}{Lemma}
IP Logged
Paul_W
Newbie
*





   
Email

Posts: 4
Re: Diameter of f(D)  
« Reply #22 on: Feb 6th, 2006, 6:13pm »
Quote Quote Modify Modify

Honestly when I asked that question, I did not  
know the answer but now after reading it ump-teen times
it is themorm 1. Interesting is the obvious circle  
of iff's.
 
Hardly got this far in complex analysis but it is  
real analysis is what I mostly know.  
 
IP Logged
shiba_san
Newbie
*





  alek_thiery  


Gender: male
Posts: 2
Re: Diameter of f(D)  
« Reply #23 on: Aug 28th, 2006, 3:14am »
Quote Quote Modify Modify

a way to solve this one when the function is injective is to compute exactly the area of f(D) in terms of the a_i's, where
f(z) = a_0 + a_1 z + a_2 z^2 + ...
 
This is not hard, just a change of variable and notice that the determinant of jacobian matrix of f at the point z is precisely |f'(z)|^2. The conclusion immediatly follows if one uses the fact that a disk is the shape with minimal diameter with a given area.
 
Nevertheless it does not works that well if f is not supposed injective, but I found the idea was nice ..
IP Logged
shiba_san
Newbie
*





  alek_thiery  


Gender: male
Posts: 2
Re: Diameter of f(D)  
« Reply #24 on: Aug 28th, 2006, 3:30am »
Quote Quote Modify Modify

Suppose that d is the diameter of f(D). Consider g(z)=f(z)-a, where a is a constant chosen such that g(D) is included in the disk of diameter d, centered at the origin. If one calls C the unit circle and because |g(z)| <= d/2 on the unit circle we get that:
 
|f'(0)| = |g'(0)| = 1/(2 pi) |int_C g(z)/z^2 dz|  
<= 1/(2 pi) 2 pi (d/2) = d/2
 
Hence d >= 2 |f'(0)| ??
IP Logged
Pages: 1 2  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board